Question

In the given figure, side BC of ∆ABC is produced to D. If ∠ACD = 128° and ∠ABC = 43°, find ∠BAC and ∠ACB.

Answer 1

Side BC of triangle ABC is produced to D.
$$\begin{gathered} \therefore \angle ACD=\angle A+\angle B[\mathrm{Exterior}\mathrm{angle}\mathrm{property}] \\ \Rightarrow 128°=\angle A+43° \\ \Rightarrow \angle A=\left(128-43\right)° \\ \Rightarrow \angle A=85° \\ \Rightarrow \angle BAC=85° \end{gathered}$$
Also, in triangle ABC,
$$\begin{gathered} \angle BAC+\angle ABC+\angle ACB=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 85°+43°+\angle ACB=180° \\ \Rightarrow 128°+\angle ACB=180° \\ \Rightarrow \angle ACB=52° \end{gathered}$$